Rational & Irrational Numbers

$\frac{p}{q} = 0$ is a rational number. If,

p and q are integers and q ≠ 0.

Hence, Option 3 is the correct option.

Every non-terminating decimal number is a real number.

Hence, Option 2 is the correct option.

7.478478 is a recurring decimal.

Hence, Option 2 is the correct option.

On solving,

$\frac{71}{75}$ = 0.94666.......

∴ $\frac{71}{75}$ is a recurring or periodic decimal.

Hence, Option 3 is the correct option.

On solving,

$\frac{13}{85} = 0.152941$........

$\frac{51}{405} = 0.1259259$......

$\frac{9}{524}$ = 0.0171755......

None of the these fraction is terminating.

Hence, Option 4 is the correct option.

(i) False, as zero is a whole number but not a natural number.

(ii) True

(iii) True

(iv) False, as $\frac{2}{5}$ is a rational number but not a whole number.

L.C.M. of 9, 12, 3 and 18 is 36.

So, converting denominator of each fraction $-\frac{5}{9}, \frac{7}{12}, -\frac{2}{3} \text{ and } \frac{11}{18}$ into 36.

$$\begin{aligned}\Rightarrow -\frac{5}{9} \times \frac{4}{4} = -\frac{20}{36} \\[1em] \Rightarrow \frac{7}{12} \times \frac{3}{3} = \frac{21}{36} \\[1em] \Rightarrow -\frac{2}{3} \times \frac{12}{12} = -\frac{24}{36} \\[1em] \Rightarrow \frac{11}{18} \times \frac{2}{2} = \frac{22}{36}.\end{aligned}$$

Since, -24 < -20 < 21 < 22

∴ -$\frac{24}{36} \lt -\frac{20}{36} \lt \frac{21}{36} \lt \frac{22}{36}$

$\Rightarrow -\frac{2}{3} \lt -\frac{5}{9} \lt \frac{7}{12} \lt \frac{11}{18}$

Difference between largest and smallest fraction :

$$\begin{aligned}\Rightarrow \frac{11}{18} - \Big(-\frac{2}{3}\Big) \\[1em] \Rightarrow \frac{11}{18} + \frac{2}{3} \\[1em] \Rightarrow \frac{11}{18} + \frac{12}{18} \\[1em] \Rightarrow \frac{23}{18} \\[1em] \Rightarrow 1.3\end{aligned}$$

Hence, fractions in ascending order are $-\frac{2}{3} \lt -\frac{5}{9} \lt \frac{7}{12} \lt \frac{11}{18}$ and required difference = 1.3

L.C.M. of 4, 8, 16 and 32 is 32.

So, converting denominator of each fraction $\frac{5}{8}, -\frac{3}{16}, -\frac{1}{4} \text{ and } \frac{17}{32}$ into 32.

$$\begin{aligned}\Rightarrow \frac{5}{8} \times \frac{4}{4} = \frac{20}{32} \\[1em] \Rightarrow -\frac{3}{16} \times \frac{2}{2} = -\frac{6}{32} \\[1em] \Rightarrow -\frac{1}{4} \times \frac{8}{8} = -\frac{8}{32} \\[1em] \Rightarrow \frac{17}{32} \times \frac{1}{1} = \frac{17}{32}.\end{aligned}$$

Since, -8 < -6 < 17 < 20.

$\therefore -\frac{8}{32} \lt -\frac{6}{32} \lt \frac{17}{32} \lt \frac{20}{32}$

$\therefore -\frac{1}{4} \lt -\frac{3}{16} \lt \frac{17}{32} \lt \frac{5}{8}$

So, in descending order.

$\Rightarrow \frac{5}{8} \gt \frac{17}{32} \gt -\frac{3}{16} \gt -\frac{1}{4}$

Sum of largest and lowest :

$$\begin{aligned}= \frac{5}{8} + \Big(-\frac{1}{4}\Big) \\[1em] = \frac{5}{8} - \frac{1}{4} \\[1em] = \frac{5 \times 4}{32} - \frac{1 \times 8}{32} \\[1em] = \frac{20 - 8}{32} \\[1em] = \frac{12}{32} \\[1em] = \frac{3}{8} \\[1em] = 0.38\end{aligned}$$

Hence, fractions in descending order are : $\frac{5}{8} \gt \frac{17}{32} \gt -\frac{3}{16} \gt -\frac{1}{4}$ and required sum = 0.38

In rational numbers, if the denominator of the fraction can be expressed in the form of 2^m × 5ⁿ, then it is a terminating decimal.

(i) $\frac{7}{16} = \frac{7}{2^4}$

So, 16 can be expressed as 2⁴ × 5⁰.

Hence, it is a terminating decimal number.

(ii) $\frac{23}{125} = \frac{23}{5^3}$

So, 125 can be expressed as 2⁰ × 5³.

Hence, it is a terminating decimal number.

(iii) $\frac{9}{14} = \frac{9}{2 \times 7}$

So, 14 cannot be expressed in form of 2^m × 5ⁿ.

Hence, it is not a terminating decimal number.

(iv) $\frac{32}{45} = \frac{32}{3^2 \times 5}$

So, 45 cannot be expressed in form of 2^m × 5ⁿ.

Hence, it is not a terminating decimal number.

(v) $\frac{43}{50} = \frac{43}{2 \times 5^2}$

So, 50 can be expressed in form of 2¹ × 5².

Hence, it is a terminating decimal number.

Negative of an irrational number is also an irrational number.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sqrt{8}(\sqrt{8} - 1) \\[1em] \Rightarrow 8 - \sqrt{8} \\[1em] \Rightarrow 8 - 2.82... \\[1em] \Rightarrow 5.17....\end{aligned}$$

which is an irrational number.

Hence, Option 2 is the correct option.

By pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ OA² = AB² + OB²

⇒ OA² = 2² + 1²

⇒ OA² = 4 + 1

⇒ OA = 5

⇒ OA = $\sqrt{5}$.

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow 2\sqrt{3} \times 3\sqrt{8} \\[1em] \Rightarrow 6\sqrt{24} \\[1em] \Rightarrow 6 \times 2\sqrt{6} \\[1em] \Rightarrow 12 \times \sqrt{6} \\[1em] \Rightarrow 12 \times 2.449..... \\[1em] \Rightarrow 29.393.......\end{aligned}$$

which is an irrational number.

Hence, Option 2 is the correct option.

$\sqrt{65}$ = 8.0622.... and $\sqrt{120}$ = 10.954......

Since, the above two nos. are non-terminating as well as non-recurring.

∴ They are irrational and in between 8 and 11.

Hence, Option 1 is the correct option.

(i) Given,

$$\begin{aligned}\Rightarrow (2 + \sqrt{2})^2 \\[1em] \Rightarrow 4 + 2 + 4\sqrt{2} \\[1em] \Rightarrow 6 + 4\sqrt{2} \\[1em] \Rightarrow 6 + 5.658.... \\[1em] \Rightarrow 11.658....\end{aligned}$$

which is irrational.

Hence, $(2 + \sqrt{2})^2$ is not a rational number.

(ii) Given,

$$\begin{aligned}\Rightarrow (3 - \sqrt{3})^2 \\[1em] \Rightarrow 9 + 3 - 6\sqrt{3} \\[1em] \Rightarrow 12 - 6\sqrt{3} \\[1em] \Rightarrow 12 - 10.392..... \\[1em] \Rightarrow 1.607......\end{aligned}$$

which is irrational.

Hence, $(3 - \sqrt{3})^2$ is not a rational number.

(iii) Given,

$$\begin{aligned}\Rightarrow (5 + \sqrt{5})(5 -\sqrt{5}) \\[1em] \Rightarrow 25 - 5\sqrt{5} + 5\sqrt{5} - 5 \\[1em] \Rightarrow 20\end{aligned}$$

which is rational.

Hence, $(5 + \sqrt{5})(5 -\sqrt{5})$ is a rational number.

(iv) Given,

$$\begin{aligned}\Rightarrow (\sqrt{3} - \sqrt{2})^2 \\[1em] \Rightarrow 3 + 2 - 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 5 - 2\sqrt{6} \\[1em] \Rightarrow 5 - 2 \times 2.449... \\[1em] \Rightarrow 5 - 4.898..... \\[1em] \Rightarrow 0.101.....\end{aligned}$$

which is irrational.

Hence, $(\sqrt{3} - \sqrt{2})^2$ is not a rational number.

(i) Squaring,

$$\begin{aligned}\Rightarrow \Big(\frac{3\sqrt{5}}{5}\Big)^2 \\[1em] \Rightarrow \frac{3\sqrt{5} \times 3\sqrt{5}}{5 \times 5} \\[1em] \Rightarrow \frac{45}{25} \\[1em] \Rightarrow \frac{9}{5} \\[1em] \Rightarrow 1\frac{4}{5}.\end{aligned}$$

Hence, square of $\frac{3\sqrt{5}}{5} = 1\frac{4}{5}$.

(ii) Squaring,

$$\begin{aligned}\Rightarrow (\sqrt{3} + \sqrt{2})^2 \\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 3 + 2 + 2\sqrt{6} \\[1em] \Rightarrow 5 + 2\sqrt{6}.\end{aligned}$$

Hence, square of $\sqrt{3} + \sqrt{2} = 5 + 2\sqrt{6}$.

(iii) Squaring,

$$\begin{aligned}\Rightarrow (\sqrt{5} - 2)^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (2)^2 - 2 \times \sqrt{5} \times 2 \\[1em] \Rightarrow 5 + 4 - 4\sqrt{5} \\[1em] \Rightarrow 9 - 4\sqrt{5}.\end{aligned}$$

Hence, square of $\sqrt{5} - 2 = 9 - 4\sqrt{5}$.

(iv) Squaring,

$$\begin{aligned}\Rightarrow (3 + 2\sqrt{5})^2 \\[1em] \Rightarrow (3)^2 + (2\sqrt{5})^2 + 2 \times 3 \times 2\sqrt{5} \\[1em] \Rightarrow 9 + 20 + 12\sqrt{5} \\[1em] \Rightarrow 29 + 12\sqrt{5}.\end{aligned}$$

Hence, square of $3 + 2\sqrt{5} = 29 + 12\sqrt{5}$.

(i) $\sqrt{2}$ = 1.41, $\sqrt{3}$ = 1.732 and $\sqrt{5}$ = 2.24

$\sqrt{2} + \sqrt{3}$ = 3.14, which is not equal to 2.24.

$\therefore \sqrt{2} + \sqrt{3} \ne \sqrt{5}$

Hence, above statement is false.

(ii) Given,

$2\sqrt{4} + 2 = 6$

Solving, L.H.S. :

$$\begin{aligned}\Rightarrow 2\sqrt{4} + 2 \\[1em] \Rightarrow 2 \times 2 + 2 \\[1em] \Rightarrow 6.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, above statement is true.

(iii) Given,

$3\sqrt{7} - 2\sqrt{7} = \sqrt{7}$

Solving L.H.S. :

$$\begin{aligned}\Rightarrow 3\sqrt{7} - 2\sqrt{7} \\[1em] \Rightarrow \sqrt{7}(3 - 2) \\[1em] \Rightarrow \sqrt{7} \times 1 \\[1em] \Rightarrow \sqrt{7}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, above statement is true.

(iv) Since, in $\frac{2}{7}$ denominator is not equal to zero.

2 and 7 have no common factor.

∴ $\frac{2}{7}$ is rational number.

Hence, above statement is false.

(v) Since, in $\frac{5}{11}$ denominator is not equal to zero.

5 and 11 have no common factor.

∴ $\frac{5}{11}$ is rational number.

Hence, above statement is true.

(vi) Both, rational and irrational numbers are real numbers.

Hence, above statement is true.

(vii) Real numbers are both rational as well as irrational number.

Hence, above statement is false.

(viii) Some rational numbers are also real numbers.

Hence, above statement is true.

(i) We need to find the set of rational numbers.

Rational numbers are :

1. Of form $\frac{p}{q}$, where q ≠ 0.

2. Integers as well as terminating and recurring decimals are rational numbers.

From the universal set

Set of rational numbers

= ${-6, -5\frac{3}{4}, -\sqrt{4}, -\frac{3}{5}, -\frac{3}{8}, 0, \frac{4}{5}, 1, 1\frac{2}{3}, 3.01, 8.47}$

(ii) Since,

$\sqrt{8} = 2.82.... \text{ and π} = 3.142....$

Since, the above numbers are neither terminating nor recurring, hence they are irrational.

From the universal set

Set of irrational numbers = ${\sqrt{8}, π}$

(iii) From the universal set

Set of integers = ${-6, -\sqrt{4}, 0, 1}$

(iv) From the universal set

Set of non-negative integers = {0, 1}.

(i) Let us assume $\sqrt{3} + \sqrt{2}$ is a rational number.

Let $\sqrt{3} + \sqrt{2}$ = x

Squaring both sides, we get;

$$\begin{aligned}\Rightarrow (\sqrt{3} + \sqrt{2})^2 = x^2\\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} = x^2\\[1em] \Rightarrow 3 + 2 + 2\sqrt{6} = x^2\\[1em] \Rightarrow 5 + 2\sqrt{6} = x^2\\[1em] \Rightarrow 2\sqrt{6} = x^2 - 5 \\[1em] \Rightarrow \sqrt{6} = \frac{x^2 - 5}{2}\end{aligned}$$

Here, x is rational,

∴ x² is rational ..................(1)

⇒ x² - 5 is rational

So, $\frac{x^2 - 5}{2}$ is rational.

But $\sqrt{6}$ is irrational, as it is square root of non-perfect square.

⇒ $\frac{x^2 - 5}{2}$ is irrational i.e. x² - 5 is irrational and so x² is irrational ....................(2)

From (1), x² is rational, and

From (2), x² is irrational

∴ We arrive at a contradiction.

So, our assumption that $\sqrt{3} + \sqrt{2}$ is a rational number is wrong.

Hence, $\sqrt{3} + \sqrt{2}$ is an irrational number.

(ii) Let us assume 3 - $\sqrt{2}$ is a rational number.

Let, 3 - $\sqrt{2}$ = x

Squaring both sides, we get;

$$\begin{aligned}\Rightarrow (3 - \sqrt{2})^2 = x^2 \\[1em] \Rightarrow (3)^2 + (\sqrt{2})^2 - 2 \times 3 \times \sqrt{2} = x^2 \\[1em] \Rightarrow 9 + 2 - 6\sqrt{2} = x^2 \\[1em] \Rightarrow 11 - 6\sqrt{2} = x^2 \\[1em] \Rightarrow 6\sqrt{2} = 11 - x^2 \\[1em] \Rightarrow \sqrt{2} = \frac{11 - x^2}{6}\end{aligned}$$

Here, x is rational,

∴ x² is rational ..................(1)

⇒ 11 - x² is rational

So, $\frac{11 - x^2}{6}$ is rational.

But $\sqrt{2}$ is irrational, as it is a square root of non-perfect square.

⇒ $\frac{11 - x^2}{6}$ is irrational i.e. 11 - x² is irrational and so x² is irrational ....................(2)

From (1), x² is rational, and

From (2), x² is irrational

∴ We arrive at a contradiction.

So, our assumption that 3 - $\sqrt{2}$ is a rational number is wrong.

Hence, 3 - $\sqrt{2}$ is an irrational number.

Let $\sqrt{3} + 2 \text{ and } \sqrt{2} - 3$ be two irrational numbers.

Sum of numbers

$$\begin{aligned}= \sqrt{3} + 2 + \sqrt{2} - 3 \\[1em] = \sqrt{3} + \sqrt{2} - 1.\end{aligned}$$

$\sqrt{3} + \sqrt{2} - 1$ is an irrational number.

Hence, required pair = $\sqrt{3} + 2 \text{ and } \sqrt{2} - 3$.

Let $\sqrt{3} + 2 \text{ and } 5 - \sqrt{3}$ be two irrational numbers.

Sum of numbers

$$\begin{aligned}= \sqrt{3} + 2 + 5 - \sqrt{3} \\[1em] = 7.\end{aligned}$$

7 is a rational number.

Hence, required pair = $\sqrt{3} + 2 \text{ and } 5 - \sqrt{3}$.

Let $\sqrt{5} + 5 \text{ and } \sqrt{2} + 5$ be two irrational numbers.

Difference of numbers

$$\begin{aligned}= \sqrt{5} + 5 - (\sqrt{2} + 5) \\[1em] = \sqrt{5} - \sqrt{2} + 5 - 5 \\[1em] = \sqrt{5} - \sqrt{2}.\end{aligned}$$

$\sqrt{5} - \sqrt{2}$ is an irrational number.

Hence, required pair = $\sqrt{5} + 5 \text{ and } \sqrt{2} + 5$.

Let $\sqrt{3} + 5 \text{ and } \sqrt{3} + 2$ be two irrational numbers.

Difference of numbers

$$\begin{aligned}= \sqrt{3} + 5 - (\sqrt{3} + 2) \\[1em] = \sqrt{3} - \sqrt{3} + 5 - 2 \\[1em] = 3.\end{aligned}$$

3 is a rational number.

Hence, required pair = $\sqrt{3} + 5 \text{ and } \sqrt{3} + 2$.

Let $\sqrt{2} \text{ and } \sqrt{3}$ be two irrational numbers.

Product of numbers

$$\begin{aligned}= \sqrt{2} \times \sqrt{3} \\[1em] = \sqrt{6}.\end{aligned}$$

$\sqrt{6}$ is an irrational number.

Hence, required pair = $\sqrt{2} \text{ and } \sqrt{3}$.

Let $5 + \sqrt{2} \text{ and } 5 - \sqrt{2}$ be two irrational numbers.

Product of numbers

$$\begin{aligned}= (5 + \sqrt{2})(5 - \sqrt{2}) \\[1em] = 25 - 5\sqrt{2} + 5\sqrt{2} - 2 \\[1em] = 23.\end{aligned}$$

23 is a rational number.

Hence, required pair = $5 + \sqrt{2} \text{ and } 5 - \sqrt{2}$.

Given,

x = $\sqrt{5} - 2$

$\frac{1}{x} = \frac{1}{\sqrt{5} - 2}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] = \frac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} \\[1em] = \frac{\sqrt{5} + 2}{5 - 4} \\[1em] = \frac{\sqrt{5} + 2}{1} \\[1em] = \sqrt{5} + 2. \\[1em] \Rightarrow x + \frac{1}{x} = \sqrt{5} - 2 + \sqrt{5} + 2 \\[1em] = 2\sqrt{5}.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

x = $1 + \sqrt{2}$

$\frac{1}{x} = \frac{1}{1 + \sqrt{2}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} \\[1em] = \frac{1 - \sqrt{2}}{(1)^2 - (\sqrt{2})^2} \\[1em] = \frac{1 - \sqrt{2}}{1 - 2} \\[1em] = \frac{1 - \sqrt{2}}{-1} \\[1em] = -1 + \sqrt{2}. \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = [1 + \sqrt{2} + (-1 + \sqrt{2})]^2 \\[1em] = [1 - 1 + \sqrt{2} + \sqrt{2}]^2 \\[1em] = [2\sqrt{2}]^2 \\[1em] = 8.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}} \\[1em] \Rightarrow \frac{2 \times 3\sqrt{3} + 3 \times 2\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow \frac{6\sqrt{3} + 6\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow \frac{12\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (\sqrt{5} - \sqrt{3})^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{3})^2 - 2 \times \sqrt{5} \times \sqrt{3} \\[1em] \Rightarrow 5 + 3 - 2\sqrt{15} \\[1em] \Rightarrow 8 - 2\sqrt{15}.\end{aligned}$$

Hence, Option 4 is the correct option.

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{3}{4 + \sqrt{7}} \times \frac{4 - \sqrt{7}}{4 - \sqrt{7}} \\[1em] \Rightarrow \frac{3(4 - \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \frac{3(4 - \sqrt{7})}{16 - 7} \\[1em] \Rightarrow \frac{3(4 - \sqrt{7})}{9} \\[1em] \Rightarrow \frac{1}{3}(4 - \sqrt{7}).\end{aligned}$$

Hence, Option 1 is the correct option.

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{7 - \sqrt{5}} \times \frac{7 + \sqrt{5}}{7 + \sqrt{5}} \\[1em] \Rightarrow \frac{7 + \sqrt{5}}{(7)^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \frac{7 + \sqrt{5}}{49 - 5} \\[1em] \Rightarrow \frac{1}{44}(7 + \sqrt{5}).\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

x = $\sqrt{2} - 1$

$\frac{1}{x} = \frac{1}{\sqrt{2} - 1}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[1em] = \frac{\sqrt{2} + 1}{(\sqrt{2})^2 - (1)^2} \\[1em] = \frac{\sqrt{2} + 1}{2 - 1} \\[1em] = \frac{\sqrt{2} + 1}{1} \\[1em] = \sqrt{2} + 1. \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = [\sqrt{2} - 1 - (\sqrt{2} + 1)]^2 \\[1em] = [\sqrt{2} - \sqrt{2} - 1 - 1]^2 \\[1em] = [-2]^2 \\[1em] = 4.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{5 - \sqrt{7}}{5 + \sqrt{7}} - \frac{5 + \sqrt{7}}{5 - \sqrt{7}} \\[1em] \Rightarrow \frac{(5 - \sqrt{7})^2 - (5 + \sqrt{7})^2}{(5 + \sqrt{7})(5 - \sqrt{7})} \\[1em] \Rightarrow \frac{(5)^2 + (\sqrt{7})^2 - 2 \times 5 \times \sqrt{7} - [(5)^2 + (\sqrt{7})^2 + 2 \times 5 \times \sqrt{7}]}{5^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \frac{25 + 7 - 10\sqrt{7} - [25 + 7 + 10\sqrt{7}]}{25 - 7} \\[1em] \Rightarrow \frac{25 - 25 + 7 - 7 - 10\sqrt{7} - 10\sqrt{7}}{18} \\[1em] \Rightarrow \frac{-20\sqrt{7}}{18} \\[1em] \Rightarrow \frac{-10\sqrt{7}}{9}.\end{aligned}$$

Hence, Option 4 is the correct option.

(i) Given,

$\sqrt{180} = \sqrt{2 \times 2 \times 3 \times 3 \times 5} = 2 \times 3 \times \sqrt{5} = 6\sqrt{5}$, which is an irrational number.

Since, 180 is a rational number and $\sqrt{180}$ is an irrational number.

Hence, $\sqrt{180}$ is a surd.

(ii) Given,

$\sqrt[4]{27} = \sqrt[4]{3^3} = (3)^{\frac{3}{4}}$, which is irrational.

Since, 27 is a rational number and $\sqrt[4]{27}$ is an irrational number.

Hence, $\sqrt[4]{27}$ is a surd.

(iii) Given,

$$\begin{aligned}\sqrt[5]{128} = \sqrt[5]{2^7} = \sqrt[5]{2^5 \times 2^2} \\[1em] = (2^5)^{\frac{1}{5}} \times (2^2)^{\frac{1}{5}} \\[1em] = 2 \times (2)^{\frac{2}{5}}\end{aligned}$$

which is irrational.

Since, 128 is a rational number and $\sqrt[5]{128}$ is an irrational number.

Hence, $\sqrt[5]{128}$ is a surd.

(iv) Given,

$\sqrt[3]{64} = \sqrt[3]{4^3} = 4^{3 \times \frac{1}{3}}$ = 4, which is rational.

Since, 64 is rational and $\sqrt[3]{64}$ is also rational.

Hence, $\sqrt[3]{64}$ is not a surd.

(v) Given,

$\sqrt[3]{25}.\sqrt[3]{40} = \sqrt[3]{1000} = \sqrt[3]{10^3} = 10^{3 \times \frac{1}{3}}$ = 10, which is rational.

Since, 1000 is rational and 10 is also rational.

Hence, $\sqrt[3]{25}.\sqrt[3]{40}$ is not a surd.

(vi) Given,

$\sqrt[3]{-125} = \sqrt[3]{(-5)^3} = (-5)^{3 \times \frac{1}{3}} = -5$, which is rational,

Since, -125 is rational and -5 is also rational.

Hence, $\sqrt[3]{-125}$ is not a surd.

(vii) Given,

$\sqrt{π}$

Since, π is irrational and $\sqrt{π}$ is also irrational.

Hence, $\sqrt{π}$ is not a surd.

(viii) Given,

$\sqrt{3 + \sqrt{2}}$

Since, $3 + \sqrt{2}$ is irrational.

Hence, $\sqrt{3 + \sqrt{2}}$ is not a surd.

(i) Given,

$\Rightarrow 5\sqrt{2}$

Rationalizing,

$$\begin{aligned}\Rightarrow 5\sqrt{2} \times \sqrt{2} \\[1em] \Rightarrow 10.\end{aligned}$$

Hence, lowest rationalizing factor of $5\sqrt{2} = \sqrt{2}$.

(ii) Given,

$$\begin{aligned}\Rightarrow \sqrt{24} \\[1em] \Rightarrow 2\sqrt{6}.\end{aligned}$$

Rationalizing,

$$\begin{aligned}\Rightarrow 2\sqrt{6} \times \sqrt{6} \\[1em] \Rightarrow 12.\end{aligned}$$

Hence, lowest rationalizing factor of $\sqrt{24} = \sqrt{6}$.

(iii) Given,

$\Rightarrow \sqrt{5} - 3$

Rationalizing,

$$\begin{aligned}\Rightarrow (\sqrt{5} - 3) \times (\sqrt{5} + 3) \\[1em] \Rightarrow (\sqrt{5})^2 - 3^2 \\[1em] \Rightarrow 5 - 9 \\[1em] \Rightarrow -4.\end{aligned}$$

Hence, lowest rationalizing factor of $\sqrt{5} - 3 = \sqrt{5} + 3$.

(iv) Given,

$\Rightarrow 7 - \sqrt{7}$

Rationalizing,

$$\begin{aligned}\Rightarrow (7 - \sqrt{7}) \times (7 + \sqrt{7}) \\[1em] \Rightarrow (7)^2 - (\sqrt{7})^2 \\[1em] \Rightarrow 49 - 7 \\[1em] \Rightarrow 42.\end{aligned}$$

Hence, lowest rationalizing factor of $7 - \sqrt{7} = 7 + \sqrt{7}$.

(v) Given,

$$\begin{aligned}\Rightarrow \sqrt{18} - \sqrt{50}\\[1em] \Rightarrow 3\sqrt{2} - 5\sqrt{2}\end{aligned}$$

Rationalizing,

$$\begin{aligned}\Rightarrow (3\sqrt{2} - 5\sqrt{2}) \times (\sqrt{2}) \\[1em] \Rightarrow (6 - 10) \\[1em] \Rightarrow -4\end{aligned}$$

Hence, lowest rationalizing factor of $\sqrt{18} - \sqrt{50} = \sqrt{2}$.

(vi) Given,

$\Rightarrow \sqrt{5} - \sqrt{2}$

Rationalizing,

$$\begin{aligned}\Rightarrow (\sqrt{5} - \sqrt{2}) \times (\sqrt{5} + \sqrt{2}) \\[1em] \Rightarrow (\sqrt{5})^2 - (\sqrt{2})^2 \\[1em] \Rightarrow 5 - 2 \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, lowest rationalizing factor of $\sqrt{5} - \sqrt{2} = \sqrt{5} + \sqrt{2}$.

(vii) Given,

$\Rightarrow \sqrt{13} + 3$

Rationalizing,

$$\begin{aligned}\Rightarrow (\sqrt{13} + 3)(\sqrt{13} - 3) \\[1em] \Rightarrow (\sqrt{13})^2 - 3^2 \\[1em] \Rightarrow 13 - 9 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, lowest rationalizing factor of $\sqrt{13} + 3 = \sqrt{13} - 3$.

(i) Given,

$\frac{2\sqrt{3}}{\sqrt{5}}$

Let us rationalize the denominator,

$$\begin{aligned}\Rightarrow \frac{2\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} \\[1em] \Rightarrow \frac{2\sqrt{15}}{(\sqrt{5})^2}\\[1em] \Rightarrow \frac{2\sqrt{15}}{5}\end{aligned}$$

Hence,$\frac{2\sqrt{3}}{\sqrt{5}} = \frac{2\sqrt{15}}{5}$

(ii) Given, $\frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

Let us rationalize the denominator,

$$\begin{aligned}\Rightarrow \frac{(\sqrt{6} - \sqrt{5}) \times (\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})\times (\sqrt{6} - \sqrt{5})}\\[1em] = \frac{(\sqrt{6} - \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2}\\[1em] = \frac{(\sqrt{6})^2 + (\sqrt{5})^2 - 2 \times \sqrt{6} \times \sqrt{5}}{6 - 5}\\[1em] = \frac{6 + 5 - 2 \times \sqrt{30}}{1}\\[1em] = 11 - 2\sqrt{30}\end{aligned}$$

Hence, $\frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} = 11 - 2\sqrt{30}$.

Given,

Equation : $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}$

Rationalizing L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \frac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{2^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{4 + 3 + 4\sqrt{3}}{4 - 3} \\[1em] \Rightarrow 7 + 4\sqrt{3}.\end{aligned}$$

Comparing $7 + 4\sqrt{3} \text{ with } a + b\sqrt{3}$, we get :

a = 7 and b = 4.

Hence, a = 7 and b = 4.

Given,

Equation : $\frac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b$

Rationalizing L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\sqrt{7} - 2}{\sqrt{7} + 2} \times \frac{\sqrt{7} - 2}{\sqrt{7} - 2} \\[1em] \Rightarrow \frac{(\sqrt{7} - 2)^2}{(\sqrt{7})^2 - 2^2} \\[1em] \Rightarrow \frac{(\sqrt{7})^2 + 2^2 - 2 \times \sqrt{7} \times 2}{7 - 4} \\[1em] \Rightarrow \frac{7 + 4 - 4\sqrt{7}}{3} \\[1em] \Rightarrow \frac{11 - 4\sqrt{7}}{3} \\[1em] \Rightarrow -\frac{4\sqrt{7}}{3} + \frac{11}{3}.\end{aligned}$$

Comparing $-\frac{4\sqrt{7}}{3} + \frac{11}{3} \text{ with } a\sqrt{7} + b$, we get :

a = $-\frac{4}{3}\text{ and } b = \frac{11}{3}$.

Hence, a = $-\frac{4}{3}\text{ and } b = \frac{11}{3}$.

Given,

Equation : $\frac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}$

Rationalizing L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{3}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \frac{3(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{3\sqrt{3} + 3\sqrt{2}}{3 - 2} \\[1em] \Rightarrow \frac{3\sqrt{3} + 3\sqrt{2}}{1} \\[1em] \Rightarrow 3\sqrt{3} + 3\sqrt{2}.\end{aligned}$$

Comparing $3\sqrt{3} + 3\sqrt{2}\text{ with } a\sqrt{3} - b\sqrt{2}$, we get :

a = 3 and b = -3.

Hence, a = 3 and b = -3.

(i) Solving,

$$\begin{aligned}\Rightarrow \frac{22}{2\sqrt{3} + 1} + \frac{17}{2\sqrt{3} - 1} \\[1em] \Rightarrow \frac{22(2\sqrt{3} - 1) + 17(2\sqrt{3} + 1)}{(2\sqrt{3} + 1)(2\sqrt{3} - 1)} \\[1em] \Rightarrow \frac{44\sqrt{3} - 22 + 34\sqrt{3} + 17}{(2\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \frac{78\sqrt{3} - 5}{12 - 1} \\[1em] \Rightarrow \frac{78\sqrt{3} - 5}{11}.\end{aligned}$$

Hence, solution = $\frac{78\sqrt{3} - 5}{11}.$

Solving,

$$\begin{aligned}\Rightarrow \frac{\sqrt{2}}{\sqrt{6} - \sqrt{2}} - \frac{\sqrt{3}}{\sqrt{6} + \sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{2}(\sqrt{6} + \sqrt{2}) - \sqrt{3}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} \\[1em] \Rightarrow \frac{\sqrt{12} + 2 - \sqrt{18} + \sqrt{6}}{(\sqrt{6})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{6 - 2} \\[1em] \Rightarrow \frac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4}.\end{aligned}$$

Hence, solution = $\frac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4}$.

(i) Substituting value of x, we get :

$$\begin{aligned}\Rightarrow x^2 = \Big(\frac{\sqrt{5} - 2}{\sqrt{5} + 2}\Big)^2 \\[1em] = \Big(\frac{\sqrt{5}^2 + 2^2 - 2 \times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 + 2 \times \sqrt{5} \times 2}\Big) \\[1em] = \frac{5 + 4 - 4\sqrt{5}}{5 + 4 + 4\sqrt{5}} \\[1em] = \frac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}}.\end{aligned}$$

Rationalizing,

$$\begin{aligned}= \frac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}} \times \frac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} \\[1em] = \frac{(9 - 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \frac{9^2 + (4\sqrt{5})^2 - 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \frac{81 + 80 - 72\sqrt{5}}{1} \\[1em] = 161 - 72\sqrt{5}.\end{aligned}$$

Hence, x² = $161 - 72\sqrt{5}.$

(ii) Substituting value of y, we get :

$$\begin{aligned}\Rightarrow y^2 = \Big(\frac{\sqrt{5} + 2}{\sqrt{5} - 2}\Big)^2 \\[1em] = \frac{(\sqrt{5})^2 + 2^2 + 2\times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 - 2\times \sqrt{5} \times 2} \\[1em] = \frac{5 + 4 + 4\sqrt{5}}{5 + 4 - 4\sqrt{5}} \\[1em] = \frac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}}\end{aligned}$$

Rationalizing,

$$\begin{aligned}= \frac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}} \times \frac{9 + 4\sqrt{5}}{9 + 4\sqrt{5}} \\[1em] = \frac{(9 + 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \frac{9^2 + (4\sqrt{5})^2 + 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \frac{81 + 80 + 72\sqrt{5}}{1} \\[1em] = 161 + 72\sqrt{5}.\end{aligned}$$

Hence, y² = $161 + 72\sqrt{5}.$

(iii) Substituting values of x and y, we get :

$$\begin{aligned}\Rightarrow xy = \frac{\sqrt{5} - 2}{\sqrt{5} + 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \\[1em] = 1.\end{aligned}$$

Hence, xy = 1.

(iv) Substituting value of x², y² and xy, we get :

$$\begin{aligned}\Rightarrow x^2 + y^2 + xy = 161 - 72\sqrt{5} + 161 + 72\sqrt{5} + 1 \\[1em] = 323.\end{aligned}$$

Hence, x² + y² + xy = 323.

(i) Substituting value of m, we get :

$$\begin{aligned}\Rightarrow m^2 = \Big(\frac{1}{3 - 2\sqrt{2}}\Big)^2 \\[1em] = \frac{1^2}{(3 - 2\sqrt{2})^2} \\[1em] = \frac{1}{3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2}} \\[1em] = \frac{1}{9 + 8 - 12\sqrt{2}} \\[1em] = \frac{1}{17 - 12\sqrt{2}}\end{aligned}$$

Rationalizing,

$$\begin{aligned}= \frac{1}{17 - 12\sqrt{2}} \times \frac{17 + 12\sqrt{2}}{17 + 12\sqrt{2}} \\[1em] = \frac{17 + 12\sqrt{2}}{17^2 - (12\sqrt{2})^2} \\[1em] = \frac{17 + 12\sqrt{2}}{289 - 288} \\[1em] = \frac{17 + 12\sqrt{2}}{1} \\[1em] = 17 + 12\sqrt{2}.\end{aligned}$$

Hence, m² = $17 + 12\sqrt{2}$.

(ii) Substituting value of n, we get :

$$\begin{aligned}\Rightarrow n^2 = \Big(\frac{1}{3 + 2\sqrt{2}}\Big)^2 \\[1em] = \frac{1^2}{(3 + 2\sqrt{2})^2} \\[1em] = \frac{1}{3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2}} \\[1em] = \frac{1}{9 + 8 + 12\sqrt{2}} \\[1em] = \frac{1}{17 + 12\sqrt{2}}\end{aligned}$$

Rationalizing,

$$\begin{aligned}= \frac{1}{17 + 12\sqrt{2}} \times \frac{17 - 12\sqrt{2}}{17 - 12\sqrt{2}} \\[1em] = \frac{17 - 12\sqrt{2}}{17^2 - (12\sqrt{2})^2} \\[1em] = \frac{17 - 12\sqrt{2}}{289 - 288} \\[1em] = \frac{17 - 12\sqrt{2}}{1} \\[1em] = 17 - 12\sqrt{2}.\end{aligned}$$

Hence, n² = $17 - 12\sqrt{2}$.

(iii) Substituting value of m and n, we get :

$$\begin{aligned}\Rightarrow mn = \frac{1}{3 - 2\sqrt{2}} \times \frac{1}{3 + 2\sqrt{2}} \\[1em] = \frac{1}{3^2 + 3 \times 2\sqrt{2} - 2\sqrt{2} \times 3 - (2\sqrt{2})^2} \\[1em] = \frac{1}{9 + 6\sqrt{2} - 6\sqrt{2} - 8} \\[1em] = \frac{1}{9 - 8} \\[1em] = \frac{1}{1} \\[1em] = 1.\end{aligned}$$

Hence, mn = 1.

(i) Substituting value of x, we get :

$\Rightarrow \frac{1}{x} = \frac{1}{2\sqrt{3} + 2\sqrt{2}}$

Rationalizing,

$$\begin{aligned}= \frac{1}{2\sqrt{3} + 2\sqrt{2}} \times \frac{2\sqrt{3} - 2\sqrt{2}}{2\sqrt{3} - 2\sqrt{2}} \\[1em] = \frac{2\sqrt{3} - 2\sqrt{2}}{(2\sqrt{3})^2 - (2\sqrt{2})^2} \\[1em] = \frac{2\sqrt{3} - 2\sqrt{2}}{12 - 8} \\[1em] = \frac{2(\sqrt{3} - \sqrt{2})}{4} \\[1em] = \frac{\sqrt{3} - \sqrt{2}}{2}.\end{aligned}$$

Hence, $\frac{1}{x} = \frac{\sqrt{3} - \sqrt{2}}{2}$.

(ii) Substituting values of x and $\frac{1}{x}$ we get :

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 2\sqrt{3} + 2\sqrt{2} + \frac{\sqrt{3} - \sqrt{2}}{2} \\[1em] = \frac{4\sqrt{3} + 4\sqrt{2} + \sqrt{3} -\sqrt{2}}{2} \\[1em] = \frac{5\sqrt{3} + 3\sqrt{2}}{2}.\end{aligned}$$

Hence, $x + \frac{1}{x} = \Big(\frac{5\sqrt{3} + 3\sqrt{2}}{2}\Big)$.

(iii) Substituting value of $\Big(x + \frac{1}{x}\Big)$, we get :

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = \Big(\frac{5\sqrt{3} + 3\sqrt{2}}{2}\Big)^2 \\[1em] = \frac{(5\sqrt{3})^2 + (3\sqrt{2})^2 + 2 \times 5\sqrt{3} \times 3\sqrt{2}}{2^2} \\[1em] = \frac{75 + 18 + 30\sqrt{6}}{4} \\[1em] = \frac{93 + 30\sqrt{6}}{4}.\end{aligned}$$

Hence, $\Big(x + \frac{1}{x}\Big)^2 = \frac{93 + 30\sqrt{6}}{4}$.

Given,

x = 1 - $\sqrt{2}$

$\therefore \frac{1}{x} = \frac{1}{1 - \sqrt{2}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1em] \Rightarrow \frac{1 + \sqrt{2}}{1^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{1 + \sqrt{2}}{1 - 2} \\[1em] \Rightarrow \frac{1 + \sqrt{2}}{-1} \\[1em] \Rightarrow -(1 + \sqrt{2}) \\[1em] \Rightarrow -1 - \sqrt{2}.\end{aligned}$$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^3 = [1 - \sqrt{2} - (-1 - \sqrt{2})]^3 \\[1em] = [1 + 1 - \sqrt{2} + \sqrt{2}]^3 \\[1em] = [2]^3 \\[1em] = 8.\end{aligned}$$

Hence, $\Big(x - \frac{1}{x}\Big)^3$ = 8.

Given,

x = 5 - $2\sqrt{6}$

$\therefore \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}} \times \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1em] = \frac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] = \frac{5 + 2\sqrt{6}}{25 - 24} \\[1em] = \frac{5 + 2\sqrt{6}}{1} \\[1em] = 5 + 2\sqrt{6}.\end{aligned}$$

By formula,

$x^2 + \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)^2 - 2$

Substituting values we get :

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = (5 - 2\sqrt{6} + 5 + 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2}$ = 98.

(i) Given,

$\Rightarrow \frac{1}{\sqrt{3} - \sqrt{2}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{\sqrt{3} + \sqrt{2}}{3 - 2} \\[1em] \Rightarrow \frac{\sqrt{3} + \sqrt{2}}{1} \\[1em] \Rightarrow \sqrt{3} + \sqrt{2} \\[1em] \Rightarrow 1.7 + 1.4 \\[1em] \Rightarrow 3.1\end{aligned}$$

Hence, $\frac{1}{\sqrt{3} - \sqrt{2}}$ = 3.1

(ii) Given,

$\Rightarrow \frac{1}{3 + 2\sqrt{2}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{1} \\[1em] \Rightarrow 3 - 2\sqrt{2} \\[1em] \Rightarrow 3 - 2 \times 1.4 \\[1em] \Rightarrow 3 - 2.8 \\[1em] \Rightarrow 0.2\end{aligned}$$

Hence, $\frac{1}{3 + 2\sqrt{2}}$ = 0.2

(iii) Given,

$\Rightarrow \frac{2 - \sqrt{3}}{\sqrt{3}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{2 - \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow \frac{2\sqrt{3} - 3}{3} \\[1em] \Rightarrow \frac{2 \times 1.7 - 3}{3} \\[1em] \Rightarrow \frac{3.4 - 3}{3} \\[1em] \Rightarrow \frac{0.4}{3} \\[1em] \Rightarrow 0.1\end{aligned}$$

Hence, $\frac{2 - \sqrt{3}}{\sqrt{3}}$ = 0.1

Solving,

$$\begin{aligned}\Rightarrow \frac{4 - \sqrt{5}}{4 + \sqrt{5}} + \frac{4 + \sqrt{5}}{4 - \sqrt{5}} \\[1em] \Rightarrow \frac{(4 - \sqrt{5})^2 + (4 + \sqrt{5})^2}{(4 + \sqrt{5})(4 - \sqrt{5})} \\[1em] \Rightarrow \frac{4^2 + (\sqrt{5})^2 - 2 \times 4 \times \sqrt{5} + 4^2 + (\sqrt{5})^2 + 2 \times 4 \times \sqrt{5}}{4^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \frac{16 + 5 - 8\sqrt{5} + 16 + 5 + 8\sqrt{5}}{16 - 5} \\[1em] \Rightarrow \frac{42}{11} \\[1em] \Rightarrow 3\frac{9}{11}.\end{aligned}$$

Hence, $\frac{4 - \sqrt{5}}{4 + \sqrt{5}} + \frac{4 + \sqrt{5}}{4 - \sqrt{5}} = 3\frac{9}{11}$.

Given,

x = $\frac{2 + \sqrt{5}}{2 - \sqrt{5}}$

Rationalizing,

$$\begin{aligned}\Rightarrow x = \frac{2 + \sqrt{5}}{2 - \sqrt{5}} \times \frac{2 + \sqrt{5}}{2 + \sqrt{5}} \\[1em] = \frac{(2 + \sqrt{5})^2}{2^2 - (\sqrt{5})^2} \\[1em] = \frac{2^2 + (\sqrt{5})^2 + 2\times 2 \times \sqrt{5}}{4 - 5} \\[1em] = \frac{4 + 5 + 4\sqrt{5}}{-1} \\[1em] = -(9 + 4\sqrt{5}). \\[1em] \Rightarrow x^2 = [-(9 + 4\sqrt{5})]^2 \\[1em] = 9^2 + (4\sqrt{5})^2 + 2\times 9 \times 4\sqrt{5} \\[1em] = 81 + 80 + 72\sqrt{5} \\[1em] = 161 + 72\sqrt{5}.\end{aligned}$$

Given,

y = $\frac{2 - \sqrt{5}}{2 + \sqrt{5}}$

Rationalizing,

$$\begin{aligned}\Rightarrow y = \frac{2 - \sqrt{5}}{2 + \sqrt{5}} \times \frac{2 - \sqrt{5}}{2 - \sqrt{5}} \\[1em] = \frac{(2 - \sqrt{5})^2}{2^2 - (\sqrt{5})^2} \\[1em] = \frac{2^2 + (\sqrt{5})^2 - 2\times 2 \times \sqrt{5}}{4 - 5} \\[1em] = \frac{4 + 5 - 4\sqrt{5}}{-1} \\[1em] = -(9 - 4\sqrt{5}). \\[1em] \Rightarrow y^2 = [-(9 - 4\sqrt{5})]^2 \\[1em] = 9^2 + (4\sqrt{5})^2 - 2\times 9 \times 4\sqrt{5} \\[1em] = 81 + 80 - 72\sqrt{5} \\[1em] = 161 - 72\sqrt{5}.\end{aligned}$$

Substituting values of x² and y², we get :

$$\begin{aligned}\Rightarrow x^2 - y^2 = (161 + 72\sqrt{5}) - (161 - 72\sqrt{5}) \\[1em] = 161 - 161 + 72\sqrt{5} + 72\sqrt{5} \\[1em] = 144\sqrt{5}.\end{aligned}$$

Hence, x² - y² = $144\sqrt{5}$.

A fraction has a terminating decimal if, when simplified, its denominator consists solely of the prime factors 2 and/or 5.

90 = 2 x 3 x 3 x 5

Since, 90 consists 2, 3 and 5.

So, $\frac{23}{90}$ is not a terminating decimal.

Hence, option 1 is the correct option.

As we know that

$$\begin{aligned}\Rightarrow \sqrt{27} \times \sqrt{3}\\[1em] = \sqrt{81}\\[1em] = 9.\end{aligned}$$

9 is rational number. So, $\sqrt{27} \times \sqrt{3}$ is also a rational number.

Hence, option 3 is the correct option.

Given, x = $\sqrt{6} - \sqrt{5}$

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{\sqrt{6} - \sqrt{5}}\\[1em] = \frac{1}{\sqrt{6} - \sqrt{5}} \times \frac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} + \sqrt{5})}\\[1em] = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2}\\[1em] = \frac{\sqrt{6} + \sqrt{5}}{6 - 5}\\[1em] = \frac{\sqrt{6} + \sqrt{5}}{1}\end{aligned}$$

Now,

$$\begin{aligned}\Rightarrow x - \frac{1}{x} = \sqrt{6} - \sqrt{5} - \Big(\frac{\sqrt{6} + \sqrt{5}}{1}\Big)\\[1em] = \sqrt{6} - \sqrt{5} - \sqrt{6} - \sqrt{5} \\[1em] = -2\sqrt{5}.\end{aligned}$$

Hence, option 4 is correct option.

Given, $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} - \frac{2 - \sqrt{3}}{2 + \sqrt{3}}$

Solving,

$$\begin{aligned}\Rightarrow \frac{(2 + \sqrt{3}) \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})} - \frac{(2 - \sqrt{3})\times (2 - \sqrt{3})}{(2 + \sqrt{3}) \times (2 - \sqrt{3})}\\[1em] \Rightarrow \frac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2} - \frac{(2 - \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}\\[1em] \Rightarrow \frac{(2)^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{4 - 3} - \frac{(2)^2 + (\sqrt{3})^2 - 2 \times 2 \times \sqrt{3}}{4 - 3}\\[1em] \Rightarrow \frac{4 + 3 + 4\sqrt{3}}{1} - \frac{4 + 3 - 4\sqrt{3}}{1}\\[1em] \Rightarrow 7 + 4\sqrt{3} - (7 - 4\sqrt{3})\\[1em] \Rightarrow 7 + 4\sqrt{3} - 7 + 4\sqrt{3}\\[1em] \Rightarrow 8\sqrt{3}.\end{aligned}$$

Hence, option 4 is correct option.

Given, a = $3\sqrt{3}$ and b = $\frac{5}{\sqrt{12}}$

$$\begin{aligned}\Rightarrow a \times b = 3 \sqrt{3} \times \frac{5}{\sqrt{12}}\\[1em] = 3 \sqrt{3} \times \frac{5}{\sqrt{4 \times 3}}\\[1em] = 3 \sqrt{3} \times \frac{5}{\sqrt{4} \times \sqrt{3}}\\[1em] = 3 \sqrt{3} \times \frac{5}{2\sqrt{3}}\\[1em] = \frac{5 \times 3 \sqrt{3}}{2\sqrt{3}}\\[1em] = \frac{15 \sqrt{3}}{2\sqrt{3}}\\[1em] = \frac{15}{2}\\[1em] = 7.5\end{aligned}$$

7.5 is rational number. Thus, a x b is a rational number.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is correct option.

Given, x = $\sqrt{5}$ + 2

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{\sqrt{5} + 2}\\[1em] = \frac{1}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2} \\[1em] = \frac{\sqrt{5} - 2}{(\sqrt{5})^2 - (2)^2}\\[1em] = \frac{\sqrt{5} - 2}{5 - 4}\\[1em] = \frac{\sqrt{5} - 2}{1}\\[1em] = \sqrt{5} - 2.\end{aligned}$$

So, statement 2 is true.

$$\begin{aligned}\Rightarrow x - \frac{1}{x} = \sqrt{5} + 2 - (\sqrt{5} - 2)\\[1em] = \sqrt{5} + 2 - \sqrt{5} + 2\\[1em] = 4.\end{aligned}$$

So, statement 1 is true.

∴ Both statements are true.

Hence, option 1 is correct option.

Given, $\frac{1}{x} = 2 + \sqrt{3}\text{ and } x + \frac{1}{x} = 4$

And,

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 4\\[1em] = x + 2 + \sqrt{3} = 4\\[1em] = x = 4 - (2 + \sqrt{3})\\[1em] = x = 4 - 2 - \sqrt{3}\\[1em] = x = 2 - \sqrt{3}.\end{aligned}$$

So, assertion (A) is true.

If, $\frac{1}{x} = 2 + \sqrt{3}$

$$\begin{aligned}\Rightarrow x = \frac{1}{2 + \sqrt{3}}\\[1em] = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \text{ (On rationalizing)} \\[1em] = \frac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2}\\[1em] = \frac{2 - \sqrt{3}}{4 - 3}\\[1em] = \frac{2 - \sqrt{3}}{1}\\[1em] = 2 - \sqrt{3}.\end{aligned}$$

So, reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is correct option.

Though all the given values $\sqrt{22}, \sqrt{23}, \sqrt{24}, \sqrt{25}, \sqrt{26}, \sqrt{27}$ lie between $\sqrt{21}$ and $\sqrt{28}$

But $\sqrt{25}$ is not an irrational number, as $\sqrt{25}$ = 5, which is a rational number.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} ÷ \frac{\sqrt{x^2 - y^2} + x}{\sqrt{x^2 + y^2} + y} \\[1em] \Rightarrow \frac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} \times \frac{\sqrt{x^2 + y^2} + y}{\sqrt{x^2 - y^2} + x} \\[1em] \Rightarrow \frac{(\sqrt{x^2 + y^2})^2 - y^2}{x^2 - (\sqrt{x^2 - y^2})^2 } \\[1em] \Rightarrow \frac{x^2 + y^2 - y^2}{x^2 - (x^2 - y^2)} \\[1em] \Rightarrow \frac{x^2}{y^2}.\end{aligned}$$

Hence, $\frac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} ÷ \frac{\sqrt{x^2 - y^2} + x}{\sqrt{x^2 + y^2} + y} = \frac{x^2}{y^2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{5}{\sqrt{20} - \sqrt{10}} \\[1em] \Rightarrow \frac{5}{2\sqrt{5} - \sqrt{10}} \\[1em] \Rightarrow \frac{5}{2 \times 2.2 - 3.2} \\[1em] \Rightarrow \frac{5}{4.4 - 3.2} \\[1em] \Rightarrow \frac{5}{1.2} \\[1em] \Rightarrow \frac{50}{12} \\[1em] \Rightarrow \frac{25}{6} = 4.2\end{aligned}$$

Another method of solving is by rationalizing,

$$\begin{aligned}\Rightarrow \frac{5}{\sqrt{20} - \sqrt{10}} \times \frac{\sqrt{20} + \sqrt{10}}{\sqrt{20} + \sqrt{10}} \\[1em] \Rightarrow \frac{5(\sqrt{20} + \sqrt{10})}{(\sqrt{20})^2 - (\sqrt{10})^2} \\[1em] \Rightarrow \frac{5(\sqrt{20} + \sqrt{10})}{20 - 10} \\[1em] \Rightarrow \frac{5(\sqrt{20} + \sqrt{10})}{10} \\[1em] \Rightarrow \frac{(2\sqrt{5} + \sqrt{10})}{2} \\[1em] \Rightarrow \frac{2 \times 2.2 + 3.2}{2} \\[1em] \Rightarrow \frac{4.4 + 3.2}{2} \\[1em] \Rightarrow \frac{7.6}{2} \\[1em] \Rightarrow 3.8\end{aligned}$$

Hence, $\frac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} ÷ \frac{\sqrt{x^2 - y^2} + x}{\sqrt{x^2 + y^2} + y}$ = 3.8 or 4.2

(i) Given,

x = $\sqrt{3} - \sqrt{2}$

$\frac{1}{x} = \frac{1}{\sqrt{3} - \sqrt{2}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{\sqrt{3} + \sqrt{2}}{3 - 2} \\[1em] \Rightarrow \sqrt{3} + \sqrt{2} \\[1em] \therefore \frac{1}{x} = \sqrt{3} + \sqrt{2} \\[2em] \Rightarrow x + \frac{1}{x} = \sqrt{3} - \sqrt{2} + (\sqrt{3} + \sqrt{2}) \\[1em] = 2\sqrt{3}.\end{aligned}$$

Hence, $x + \frac{1}{x} = 2\sqrt{3}$.

(ii) Solving,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = (\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2 \\[1em] = (\sqrt{3})^2 + (\sqrt{2})^2 - 2 \times \sqrt{3} \times \sqrt{2} + (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 3 + 2 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} \\[1em] \Rightarrow 10.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2} = 10$.

(iii) Solving,

$$\begin{aligned}\Rightarrow x^3 + \frac{1}{x^3} = (\sqrt{3} - \sqrt{2})^3 + (\sqrt{3} + \sqrt{2})^3 \\[1em] = (\sqrt{3})^3 - (\sqrt{2})^3 - 3 \times \sqrt{3} \times \sqrt{2} \times (\sqrt{3} - \sqrt{2}) + (\sqrt{3})^3 + (\sqrt{2})^3 + 3 \times \sqrt{3} \times \sqrt{2} \times (\sqrt{3} + \sqrt{2}) \\[1em] = 3\sqrt{3} - 2\sqrt{2} - 3\sqrt{6}(\sqrt{3} - \sqrt{2}) + 3\sqrt{3} + 2\sqrt{2} + 3\sqrt{6}(\sqrt{3} + \sqrt{2}) \\[1em] = 3\sqrt{3} + 3\sqrt{3} - 2\sqrt{2} + 2\sqrt{2} - 3\sqrt{18} + 3\sqrt{12} + 3\sqrt{18} + 3\sqrt{12} \\[1em] = 6\sqrt{3} + 6\sqrt{12} \\[1em] = 6\sqrt{3} + 6 \times 2\sqrt{3} \\[1em] = 6\sqrt{3} + 12\sqrt{3} \\[1em] = 18\sqrt{3}.\end{aligned}$$

Hence, $x^3 + \frac{1}{x^3} = 18\sqrt{3}$.

(iv) Substituting values from part (i), (ii) and (iii), we get :

$$\begin{aligned}\Rightarrow x^3 + \frac{1}{x^3} - 3\Big(x^2 + \frac{1}{x^2}\Big) + x + \frac{1}{x} = 18\sqrt{3} - 3 \times 10 + 2\sqrt{3} \\[1em] = 20\sqrt{3} - 30 \\[1em] = 10(2\sqrt{3} - 3).\end{aligned}$$

Hence, $x^3 + \frac{1}{x^3} - 3\Big(x^2 + \frac{1}{x^2}\Big) + x + \frac{1}{x} = 10(2\sqrt{3} - 3)$.

(i) True

For example, $\sqrt{3}$ is irrational and $-\sqrt{3}$ is also irrational.

(ii) False

For example, 2 is rational number and $\sqrt{2}$ is irrational, product of these numbers i.e. $2\sqrt{2}$ is also irrational.

Steps of construction :

1. Draw a line segment XY.

2. Draw OB = 1 cm which is perpendicular to the line XY at point O.

3. From B draw an arc of 2 cm cutting XY at A.

4. Join BA and OA.

So, OAB is the right angle triangle.

By pythagoras theorem,

⇒ AB² = OB² + OA²

⇒ 2² = 1² + OA²

⇒ OA² = 4 - 1

⇒ OA² = 3

⇒ OA = $\sqrt{3}$ cm.

Hence, OA is the required line of $\sqrt{3}$ cm.

Steps of construction :

1. Draw a line segment XY.

2. Draw OB = 1 cm which is perpendicular to the line XY at O.

3. From B draw an arc of 3 cm cutting XY at A.

4. Join BA and OA.

So, OAB is the right angle triangle.

By pythagoras theorem,

⇒ AB² = OB² + OA²

⇒ 3² = 1² + OA²

⇒ OA² = 9 - 1

⇒ OA² = 8

⇒ OA = $\sqrt{8}$ cm.

Hence, OA is the required line of $\sqrt{8}$ cm.

(i) Given,

x = 2 + $\sqrt{3}$

$\therefore \frac{1}{x} = \frac{1}{2 + \sqrt{3}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \\[1em] \Rightarrow \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{2 - \sqrt{3}}{4 - 3} \\[1em] \Rightarrow 2 - \sqrt{3}. \\[1em] \therefore \frac{1}{x} = 2 - \sqrt{3}\end{aligned}$$

Substituting value of x and $\frac{1}{x} \text{ in } x^3 + \frac{1}{x^3}$, we get :

$$\begin{aligned}\Rightarrow x^3 + \frac{1}{x^3} = (2 + \sqrt{3})^3 + (2 -\sqrt{3})^3 \\[1em] = 2^3 + (\sqrt{3})^3 + 3 \times 2 \times \sqrt{3} \times (2 + \sqrt{3}) + 2^3 - (\sqrt{3})^3 - 3 \times 2 \times \sqrt{3} \times (2 - \sqrt{3}) \\[1em] = 8 + 3\sqrt{3} + 6\sqrt{3}(2 + \sqrt{3}) + 8 - 3\sqrt{3} - 6\sqrt{3}(2 - \sqrt{3}) \\[1em] = 8 + 8 + 3\sqrt{3} - 3\sqrt{3} + 12\sqrt{3} + 18 - 12\sqrt{3} + 18 \\[1em] = 8 + 8 + 18 + 18 \\[1em] = 52.\end{aligned}$$

Hence, proved that $x^3 + \frac{1}{x^3} = 52$.

Given,

x = $3 + 2\sqrt{2}$

$\therefore \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2} \\[1em] \therefore \frac{1}{x} = 3 - 2\sqrt{2}\end{aligned}$$

Substituting value of x and $\frac{1}{x} \text{ in } x^2 + \frac{1}{x^2}$, we get :

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = (3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2 \\[1em] \Rightarrow 3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2} + 3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2} \\[1em] \Rightarrow 9 + 8 + 12\sqrt{2} + 9 + 8 - 12\sqrt{2} \\[1em] \Rightarrow 34.\end{aligned}$$

Hence, proved that $x^2 + \frac{1}{x^2} = 34.$

Given,

Equation : $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{3}\sqrt{2}(\sqrt{3} - \sqrt{2})}{\sqrt{3}\sqrt{2}(\sqrt{3} + \sqrt{2})} + \frac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} + \frac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \frac{(\sqrt{3} - \sqrt{2})^2 + 2\sqrt{3}(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{(\sqrt{3})^2 + (\sqrt{2})^2 - 2\times \sqrt{3} \times \sqrt{2} + 6 + 2\sqrt{6}}{3 - 2} \\[1em] \Rightarrow \frac{3 + 2 - 2\sqrt{6} + 6 + 2\sqrt{6}}{1} \\[1em] \Rightarrow 11.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11.

(i) Given,

⇒ x² = 6

⇒ x = $\sqrt{6}$

⇒ x = 2.449.....

Since, x is non-terminating as well as non-recurring.

Hence, proved that x is irrational.

(ii) Given,

⇒ x² = 0.009

⇒ x = $\sqrt{0.009}$

⇒ x = 0.094.....

Since, x is non-terminating as well as non-recurring.

Hence, proved that x is irrational.

(iii) Given,

⇒ x² = 27

⇒ x = $\sqrt{27}$

⇒ x = 5.196.....

Since, x is non-terminating as well as non-recurring.

Hence, proved that x is irrational.

(i) Given,

⇒ x² = 16

⇒ x = $\sqrt{16}$

⇒ x = $\pm 4$.

Integers are considered as rational numbers.

Hence, proved that x is rational.

(ii) Given,

⇒ x² = 0.0004

⇒ x = $\sqrt{0.0004}$

⇒ x = $0.02$.

Terminating decimals are considered as rational numbers.

Hence, proved that x is rational.

(iii) Given,

⇒ x² = $1\frac{7}{9}$

⇒ x² = $\frac{16}{9}$

⇒ x = $\sqrt{\frac{16}{9}}$

⇒ x = $\frac{4}{3}$ = 1.333.....

Recurring decimals are considered as rational number.

Hence, proved that x is rational.

Rationalizing a general term,

$\frac{1}{\sqrt{n} + \sqrt{n+1}}$

To rationalize, we multiply the numerator and denominator by the conjugate, $(\sqrt{n+1} - \sqrt{n})$:

$$\begin{aligned}\frac{1}{\sqrt{n+1} + \sqrt{n}} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} \\[1em] = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} \\[1em] = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} \\[1em] = \sqrt{n+1} - \sqrt{n}\end{aligned}$$

Expanding the series,

First term: $\frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1$

Second term: $\frac{1}{\sqrt{2} + \sqrt{3}} = \sqrt{3} - \sqrt{2}$

Third term: $\frac{1}{\sqrt{3} + \sqrt{4}} = \sqrt{4} - \sqrt{3}$

...

Last term: $\frac{1}{\sqrt{2025} + \sqrt{2026}} = \sqrt{2026} - \sqrt{2025}$

Adding all the series,

$(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{2026} - \sqrt{2025})$

$\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \dots + \sqrt{2026} - \sqrt{2025}$

Everything cancels except the first and last term.

-1 + $\sqrt{2026}$.

Hence, solution = $\sqrt{2026}$ - 1.

(a) No, because rational - irrational = irrational

Example : 2 - $\sqrt{3}$ is an irrational number as $\sqrt{3}$ is an irrational number.

Hence, required answer is no.

(b) As 0.5555 is a repeating decimal number, so it is a rational number.

Hence, required answer is Yes.

(c) 0.505005 is non-terminating and non-repeating number.

So it is an irrational number, not a rational number.

Hence, required answer is No.

(d) Rationalizing factor of 3 - $\sqrt{7}$ is 3 + $\sqrt{7}$

= (3 - $\sqrt{7}$) × (3 - $\sqrt{7}$)

= (3)² - $(\sqrt{7})^2$

= 9 - 7

= 2.

2 is a rational number.

Hence, required answer is Yes.

(i) For a number of the form a + $b\sqrt{c}$, the rationalizing factor is a - $b\sqrt{c}$.

So, for 7 + $4\sqrt{3}$, the rationalizing factor is 7 - $4\sqrt{3}$.

Hence rationalizing factor = 7 - $4\sqrt{3}$.

(ii) Reciprocal of 7 + $4\sqrt{3}$ is :

$\frac{1}{7 + 4 \sqrt{3}}$

Multiplying numerator and denominator by the rationalising factor :

$\frac{1}{7 + 4 \sqrt{3}} × \Big(\frac{7 - 4 \sqrt{3}}{7 - 4 \sqrt{3}} \Big)$

= $\frac{7 - 4 \sqrt{3}}{7^2 - (4 \sqrt{3})^2}$

= $\frac{7 - 4 \sqrt{3}}{49 - 48}$

= 7 - $4\sqrt{3}$.

Hence, reciprocal of 7 + $4\sqrt{3}$ = 7 - $4\sqrt{3}$. Thus, the reciprocal and the rationalising factor both are same.

(iii) Given,

x = 7 + $4\sqrt{3}$,

$\frac{1}{x}$ = 7 - $4\sqrt{3}$

$x + \frac{1}{x} = (7 + 4 \sqrt{3}) + (7 - 4\sqrt{3})$ = 14.

Hence, value = 14.